Given two positive number N and X. The task is to find the sum of digits of a number formed by N repeating X number of times until sum become single digit.
Input : N = 24, X = 3 Output : 9 Number formed after repeating 24 three time = 242424 Sum = 2 + 4 + 2 + 4 + 2 + 4 = 18 Sum is not the single digit, so finding the sum of digits of 18, 1 + 8 = 9 Input : N = 4, X = 4 Output : 7
As discussed in this post, recursive sum of digits is 9 if number is multiple of 9, else n % 9. Since divisibility and modular arithmetic are compatible with multiplication, we simply find result for single occurrence, multiply result with x and again find the result.
How does this work ?
Lets N = 24 and X = 3.
So, sumUntilSingle(N) = 2 + 4 = 6.
Multiplying 6 by 3 = 18
sumUntilSingle(18) = 9.
Below is the implemenatation of this approach:
- Remove repeated digits in a given number
- Count ways to spell a number with repeated digits
- Recursive sum of digits of a number is prime or not
- Check if a number is magic (Recursive sum of digits is 1)
- Find M-th number whose repeated sum of digits of a number is N
- Greatest number less than equal to B that can be formed from the digits of A
- Find maximum number that can be formed using digits of a given number
- Find all strings formed from characters mapped to digits of a number
- Find the Largest Cube formed by Deleting minimum Digits from a number
- Squares of numbers with repeated single digits | Set 1 (3, 6 and 9)
- Total numbers with no repeated digits in a range
- Maximum possible time that can be formed from four digits
- Smallest multiple of N formed using the given set of digits
- Sum of all numbers that can be formed with permutations of n digits
- Count numbers formed by given two digit with sum having given digits
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Improved By : vt_m